Two circles of equal radii cut each other at P and Q, so that the centre of one lies on the other. A straight line through P cuts the circle again at A and B. Prove that `Delta`QAB is equilateral triangle.

Two circles of equal radii cut each other at P and Q, so that the cent

1.	Two circles of equal radii cut each other at P and Q, so that the centre of one lies on the other. A straight line through P cuts the circle again at A and B. Prove that `Delta`QAB is equilateral triangle.
Answer» `QC^2=r^2-(r/2)^2` =`3/4r^2` `QC=sqrt3/2r` `/_QOP=2*/_QOC` =`120^0` `/_QRP=120/2=60^0` `/_QRP and /_QBP` are inscribed on the arc PQ, `theta=60^0` `/_QOP=/_QAP=120^0` they are inscribed in the same are QSR `/_QAB=180^0=/_QAP=180^0=120^0=60^0` All the angles are `60^0` Therefore, QAB is equilateral triangle.

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Two circles of equal radii cut each other at P and Q, so that the centre of one lies on the other. A straight line through P cuts the circle again at A and B. Prove that `Delta`QAB is equilateral triangle.

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